PAT 1117 Eddington Number [难]

1117 Eddington Number （25 分）

British astronomer Eddington liked to ride a bike. It is said that in order to show off his skill, he has even defined an "Eddington number", E -- that is, the maximum integer E such that it is for E days that one rides more than E miles. Eddington's own E was 87.

Now given everyday's distances that one rides for N days, you are supposed to find the corresponding E (≤N).

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤10​5​​), the days of continuous riding. Then N non-negative integers are given in the next line, being the riding distances of everyday.

Output Specification:

For each case, print in a line the Eddington number for these N days.

Sample Input:

10
6 7 6 9 3 10 8 2 7 8

Sample Output:

6

题目大意：E是有E天骑行超过E的最大整数。

对于本题输出6，有7 9 10 8 7 8这6天是超过6的，等于6的不算，所以得出了6.

//发现自己并不会写，不会写啊！！好难啊！这个逻辑！！！

柳神的解答:

//虽然题干挺短，但是逻辑挺难的，